# A hollow verticle cylinder of radius r and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim, at point A, and given a horizontal speed u, tangential to the rim?

##
It leaves the lower rim at point B, vertically below A. If n is an integer then?

A) #u/(2pir)(sqrt((2h)/g)) = n#

B) #h/(2pir) = n#

C) #u/sqrt(2gh)=n#

It leaves the lower rim at point B, vertically below A. If n is an integer then?

A)

B)

C)

I get

(A)

Inserting various values we get

Equating RHSs of (1) and (2) we get

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The motion of the particle is subject to the law of conservation of energy and conservation of angular momentum. Initially, the particle has kinetic energy due to its speed u and potential energy due to its height above the bottom of the cylinder. As it moves downward, its potential energy decreases, and its kinetic energy increases.

The angular momentum of the particle about the center of the cylinder is conserved because there is no net torque acting on it. Therefore, the angular momentum at the top (where it starts) is equal to the angular momentum at any point as it moves down.

By applying the conservation of energy and conservation of angular momentum, you can find equations that relate the particle's speed at any point to its distance from the bottom of the cylinder. These equations will help determine the particle's motion within the cylinder.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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